我想知道如何将 map 字典的值列表设为 dataframe

Question

I would like to make this

City
Seoul
Busan
Deagu
Seoul
Seoul
Busan

into this which now has latitude and longtitue

City	Latitude	Logitude
Seoul	127	50
Busan	128	51
Daegu	129	52
Seoul	127	50
Seoul	127	50
Busan	128	51

I tried this using code below

import pandas as pd

dict1 = {"Seoul":[127,50],
         "Busan":[128,51],
         "Daegu":[129,52]}

data = {'City':['Seoul', 'Busan', 'Daegu', 'Seoul','Seoul','Busan']}

test1 = pd.DataFrame(data=data)

for city, list1 in dict1.items():
    print(city," ", list1[0],list1[1])

for num, city, list1 in zip(range(7),dict1.items()):
    if test1.loc[:,"City"] == city:
        test1.loc["Latitude"] = list1[0]
        test1.loc["Longitude"] = list1[1]

But it returned error for not having enough elements for for loop and need more efficient way better than iterating for length of list Have a great day!

Answer 1

Use DataFrame.from_dict with DataFrame.join :

df1 = pd.DataFrame.from_dict(dict1, orient='index', columns=['Latitude','Longitude'])
test1 = test1.join(df1, on='City')
print (test1)
    City  Latitude  Longitude
0  Seoul       127         50
1  Busan       128         51
2  Daegu       129         52
3  Seoul       127         50
4  Seoul       127         50
5  Busan       128         51

Answer 2

Try using the following one-liner:

print(pd.DataFrame([[k] + v for k, v in dict1.items()], columns=['City', 'Latitude', 'Longitude']).set_index('City').loc[data['City']].reset_index())

Or as @jezrael mentioned below, you could also do unpacking with:

print(pd.DataFrame([(k, *v) for k, v in dict1.items()], columns=['City', 'Latitude', 'Longitude']).set_index('City').loc[data['City']].reset_index())

Output:

    City  Latitude  Longitude
0  Seoul       127         50
1  Busan       128         51
2  Daegu       129         52
3  Seoul       127         50
4  Seoul       127         50
5  Busan       128         51